Submitted by mrmonkey on Tue, 10/17/2017 - 16:01

I don't understand how you can show that $e^xe^{-x} = 1$? I understand that because of the differentiation it must be a constant, and that when $x=0 , e^x=1$, but how does this show that $e^xe^{-x} = 1$? I am possibly just being thick, or perhaps something is assumed which I thought wasn't meant to be?

## Constant function

We have that the derivative is constant (for anyone stuck here, take $a=1$ and $b=-1$ in the first result in this part).

Hence we have $e^xe^{-x}=c$ for some $c$. Substituting $x=0$ gives:

\begin{align*}

e^0e^{-0}&=c\\

e^0e^0&=c\\

1 \times 1 &=c\\

1&=c

\end{align*}

Hence we have $c=1$ and $e^xe^{-x}=1$.